As Sandor and Martin pointed out above, it is ok if the subschemes intersect in the emptyset pairwise. Here's an example where it's false without that hypotheses, notice that the varieties are smooth and they intersect pairwise with normal crossings. EDIT: as t3suji pointed out in the comments, these varieties don't intersect transversally in the ambient space, just in the ambient $z = 0$ plane EndOfEdit Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z-1)$. Notice that $Z_3$ doesn't intersect any of the other subschemes.
Then, $I_1 \cap I_2 = (z, xy)$.
However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$.
These ideals are not equal clearly. Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal. However, just to be sure, I also did the following computation (with Macaulay2):
$$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3 - z^2, yz^2 - yz, xz^2 - xz).$$
$$I_1 \cap I_2 \cap I_3 = (z^2 -z, xz, xy).$$
Macaulay2 also confirmed that the ideals were not equal.
Ok, let me now give a proof of a correct statement.
Lemma: Suppose that subschemes $Z_1, \dots, Z_k$ have pairwise trivial intersection in some ambient Noetherian scheme $X$. Then $I_{Z_1} \dots I_{Z_K} = I_{Z_1} \cap I_{Z_k}$.
Proof: The statement is local so we may assume that $X$ is the spectrum of a local ring $(R, \mathfrak{m})$. Now, since $I_{Z_1} + I_{Z_2} = R$, at least one of those ideals must equal $R$ (if not, both would be in the maximal ideal $\mathfrak{m}$, and so would their sum). Likewise with all pairs. Therefore, at most one of the ideals $I_{Z_i}$ is not equal to $R$. But now the statement is obvious. $R \cdot R \dots I_{Z_i} \dots R = I_{Z_i} = R \cap R \cap \dots I_{Z_i} \cap \dots R$.